Two isotropic sources, equal amplitude, opposite phase

Consider two isotropic point sources 1 and 2 spaced a distance d apart on the x-axis, with the reference (origin) at the midpoint. A far point P is in direction φ measured from the array axis. Each source alone gives field E0 at P.

                    to far point P
                   /
                  / φ
   (1)-----------O-----------(2)
    |<--- d/2 --->|<--- d/2 --->|
   path from (2) shorter by (d/2)cosφ
   path from (1) longer  by (d/2)cosφ

Path difference between the two waves = d cosφ, so the phase difference due to path is

ψ = β d cosφ,      β = 2π/λ

With respect to the midpoint, wave 2 leads by ψ/2 and wave 1 lags by ψ/2.

Field for equal amplitude, opposite phase

Currents equal in magnitude but opposite in phase (180°). By superposition:

E = −E0 e^(−jψ/2) + E0 e^(+jψ/2)
  = 2jE0 sin(ψ/2)
|E| = 2E0 sin[(βd/2) cosφ]

For d = λ/2 (βd/2 = π/2), normalized: E_n = sin[(π/2) cosφ].

• Maxima: (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180° (along the axis, end-fire).
• Minima: (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270°.
• Half power: sin[(π/2)cosφ] = ±1/√2 → cosφ = ±1/2 → φ = 60°, 120°, 240°, 300° → HPBW = 120° per lobe.

            φ=90° (null)
                |
     .--.       |       .--.
 180°(    )---(1)O(2)---(    ) 0°
     '--'       |       '--'
                |
  End-fire figure-of-eight

General conditions (any d):

• Maxima: (βd/2)cosφ = ±(2n + 1)π/2 → cosφ = ±(2n + 1)λ/(2d)
• Minima: (βd/2)cosφ = ±nπ → cosφ = ±nλ/d

So the opposite-phase pair is a simple end-fire array.